求$(2x)^{4x}$的导数：计算结果与在线计算器不符的疑问

Hey there, let's walk through your derivative work for ( y = (2x)^{4x} ) — I can see where the mismatch with Symbolab is coming from, it's just a small simplification error at the end!

First, let's recap your steps (most of them are totally spot-on):

• Starting with the original equation: ( y = (2x)^{4x} )
• Taking natural log of both sides: ( \ln(y) = \ln\left((2x)^{4x}\right) )
• Applying the log power rule: ( \ln(y) = 4x \cdot \ln(2x) )
• Differentiating both sides (product rule + chain rule):
  ( \frac{1}{y} \cdot y' = 4\ln(2x) + 4x \cdot \frac{d}{dx}[\ln(2x)] )
  For the derivative of ( \ln(2x) ), you nailed the chain rule: ( \frac{d}{dx}[\ln(2x)] = \frac{1}{2x} \cdot 2 = \frac{1}{x} )
  Substituting back in: ( \frac{1}{y} \cdot y' = 4\ln(2x) + 4x \cdot \frac{1}{x} = 4\ln(2x) + 4 )

Here's the tiny slip-up: when you substituted ( y ) back into the equation, you wrote ( 2x^{4x} ) instead of ( (2x)^{4x} ). Those parentheses around ( 2x ) are critical — without them, you're only raising ( x ) to the ( 4x ) power, not the entire ( 2x ) term.

The corrected final step looks like this:
( y' = (2x)^{4x} \left( 4\ln(2x) + 4 \right) )

You can also factor out the 4 to simplify it further if you'd like:
( y' = 4(2x)^{4x} \left( \ln(2x) + 1 \right) )

This should match exactly what Symbolab outputs! If you want to dig deeper into any part of this, feel free to ask.

内容的提问来源于stack exchange，提问作者Michael Ramage MikeRamage